 AC*. by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is to the base BD (Prop. III.) ; now, the square ABHL has been proved to be equivalent...  Progressive Lessons in Applied Science - Page 97by Edward Sang - 1875Full view - About this book
 | 1860 - 462 pages
...between the sides including the right angle. When the sides of a right-angled triangle are commensurable, the perpendicular let fall from the vertex of the right angle to the hypothenuse is commensurable with the sides, because the two partial triangles thus formed are similar... | |
 | John Daniel Runkle - Mathematics - 1860 - 460 pages
...between the sides including the right angle. When the sides of a righ<>angled triangle are commensurable, the perpendicular let fall from the vertex of the right angle to the hypothenuse is commensurable with the sides, because the two partial triangles thus formed are similar... | |
 | Benjamin Greenleaf - Geometry - 1862 - 518 pages
...hypothenuse diminished by the square of the other side ; thus, A B2 is equivalent to B C2 — AC*. by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC... | |
 | Benjamin Greenleaf - Geometry - 1861 - 638 pages
...the hypothenuse diminished by the square of the other side ; thus, O2 is equivalent to B~Ca — AC". by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude BF, the square BCGF is to the rectangle BDEF as the hase BG... | |
 | Benjamin Greenleaf - Geometry - 1863 - 504 pages
...the hypothenuse diminished by the square of the other side ; thus, AB2 is equivalent to B C2 — AC*. by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC... | |
 | Benjamin Greenleaf - Geometry - 1868 - 340 pages
...the hypothenuse diminished by the square of the other side ; thus, AB is equivalent to B C" — AC. by the perpendicular let fall from the vertex of the right angle. For, on account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC... | |
 | University of the State of New York. Examination Department - Examinations - 1894 - 412 pages
...passing through two given points and having a given radius. (Give proof.) 11 In a right angled triangle, the perpendicular let fall from the vertex of the right angle to the opposite side divides that side into two segments which are 4^ ft and 5$ ft. Find the three sides of... | |
 | Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 304 pages
...equilateral triangle is three fourths the square on a side. 7. If in a right triangle a perpendicular is let fall from the vertex of the right angle to the hypotenuse, show that the areas of the two triangles thus formed are in the same ratio as the adjacent segments... | |
 | Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 300 pages
...equilateral triangle is three fourths the square on a side. 7. If in a right triangle a perpendicular is let fall from the vertex of the right angle to the hypotenuse, show that the areas of the two triangles thus formed are in the same ratio as the adjacent segments... | |
 | Rochester (N.Y.). Board of Education - Education - 1928 - 616 pages
...cut off as the other side is to the corresponding segment. If in a right triangle a perpendicular is let fall from the vertex of the right angle to the hypotenuse : 6. Either leg is a mean proportional between the hypotenuse and the adjacent segment. Group II. 1.... | |
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